Joan brings a cup of hot tea into a room and places the cup on a table - Edexcel - A-Level Maths Pure - Question 6 - 2011 - Paper 4

At time t = 5 minutes, the temperature decreases to 55°C:

$θ = 55 = 20 + 70e^{-5k}$

Rearranging this gives:

$55 - 20 = 70e^{-5k}$ $35 = 70e^{-5k}$

Dividing both sides by 70 results in:

$\frac{35}{70} = e^{-5k}$ $\frac{1}{2} = e^{-5k}$

Taking the natural logarithm of both sides:

$ln(\frac{1}{2}) = -5k$

This simplifies to:

$-ln(2) = -5k$

Thus,

$k = \frac{ln(2)}{5}$

To find the rate of temperature change, we differentiate the equation:

$θ = 20 + 70e^{-kt}$

Using the chain rule:

$\frac{dθ}{dt} = -70ke^{-kt}$

From part (b), we found that k = \frac{ln(2)}{5}, so substituting this gives:

$\frac{dθ}{dt} = -70 \cdot \frac{ln(2)}{5} e^{-\frac{ln(2)}{5}(t)}$

At t = 10:

$\frac{dθ}{dt} = -70 \cdot \frac{ln(2)}{5} e^{-\frac{ln(2)}{5}(10)}$

Calculating:

$\frac{dθ}{dt} = -14ln(2) \cdot e^{-2ln(2)}$ $= -14ln(2) \cdot \left(\frac{1}{(2)^2}\right) = -\frac{14ln(2)}{4} = -3.5ln(2)$

Evaluating numerically gives:

$\frac{dθ}{dt} ≈ -2.426013132...$

Thus, the rate at which the temperature of the tea is decreasing at that instant is approximately:

$\boxed{2.426} °C/min$ (to 3 decimal places).